of 1 79

A Precise Theory for the Proton Radius and the Elements as Mathematical Sets!

Seventh Edition!

By!

Ian Beardsley!

of 2 79

Contents!

Preface……………………………………………………….3!

Data…………………………………………………………..4!

Abstract………………………………………………………5!

Introduction………………………………………………….6!

1.0 Ground Work……………………………………………6!

2.0 The Elements As Proton Sets….. …………………..10!

3.0 Solving The Calendar………………………………….14!

4.0 Predicting The Meter…………………………………..23!

5.0 The Constant k…………………………………………26!

6.0 Discussion……………………………………………..35!

7.0 Solar System A Quantum Mechanical System…….36!

8.0 Solar Luminosity A QM State………………………..43!

9.0 Genesis Project Equations……………………………47!

10.0 Another Theory For The Radius Of A Proton……..49!

11.0 The Macro and Micro Planck Constants………….53!

12.0 The Exact Radius of a Proton………………………57!

13.0 The Central Role of the Universe in the!

Emergence of life…………………………………….61!

of 3 79

Preface

One of things I wanted to go into in this paper but didn’t is that you can speak of the structure of

the solar system even though it changes with time. This is important to understand when I refer

to sizes the Moon and the planets, and their orbital distances.

The whole object of developing a theory for the way planetary systems form is that they meet the

following criterion: They predict the Titius-Bode rule for the distribution of the planets; the

distribution gives the planetary orbital periods from Newton’s Universal Law of Gravitation. The

distribution of the planets is chiefly predicted by three factors: The inward forces of gravity from

the parent star, the outward pressure gradient from the stellar production of radiation, and the

outward inertial forces as a cloud collapses into a flat disc around the central star. These forces

separate the flat disc into rings, agglomerations of material, each ring from which a different

planet forms at its central distance from the star (it has a thickness). In a theory of planetary

formation from a primordial disc, it should predict the Titius-Bode rule for the distribution of

planets. When I speak of the state of the solar system I speak of this point toward which the

solar system has formed and not the small changes that happen over millions of years due to

mutual interference between the bodies. In fact, mutual interference has torn apart possible

forming planets resulting in the current distribution we have today, because the current

distribution is more or less stable. The asteroid belt is a good example of this — it is a location

where a planet cannot form due to harmonic (repetitive) action on the orbital period at its

distance by orbital periods of planets beyond it. In short we take the state of the solar system as

an inflection point between what it became, and what it might minutely be going away from in

billions of years after it dies and can no longer support life.

of 4 79

The natural constants measure the properties of space and time. We can list some of them here:

(Proton Mass)

(Planck Constant)

(Proton Radius)

(Gravitational Constant)

(light speed)

(Fine Structure Constant)

: 1.67262 × 10

−27

h : 6.62607 × 10

−34

J ⋅ s

: 0.833 × 10

−15

G:6.67408 × 10

−11

c : 299,792, 459m /s

α : 1/137

= q

= 1.6022E − 19coulom bs

= 8.988E 9

of 5 79

In the seventh edition we include DNA and RNA, we had the hydrocarbons, amino acids, and

ribose.

In the sixth edition since we showed our theory includes the hydrocarbons, the backbones of life

chemistry, we needed to now predict the amino acids, the building blocks of life. We do that

then we predict ribose the fundamental sugar of life that is a primary component to DNA and

RNA that provide the instructions for life. Now that we have the amino acids and ribose, we

need the purines, or nucleobases adenine, guanine, cytosine, thymine, and uracil, and the

phosphate backbone DNA and RNA. The findings here may shed light on origins of life on Earth,

the prebiotic path.

In the fourth edition in section 12.0 we suggest if the nature of space and time is such the radius

of a proton is given by the golden ratio, then we can determine its value exactly in theory and we

find this in turn gives the momentum-distance of quantum mechanics, and the asymptote of the

special relativity hyperbola of spacetime diagrams that is the world-line of a light signal.

In the third edition we create in section 11.0 a Planck constant on the macro-scale and compare

it to the Planck constant of quantum mechanics and see that the Earth/Moon/Sun system is a

quantum mechanical system as well because the kinetic energy of the Earth is an eigenvalue

mapping the macro-scale Planck constant to the micro-scale Planck constant, as well as the

Planck constants being eigenvalues as in quantum mechanics for proton/electron systems but

here for the Earth kinetic energy around the Sun. This is done by setting out to show that the

second is a natural unit, a natural constant, which we achieve.

The proton is the fundamental charged particle that makes up atoms. We develop a theory of

inertia with the proton as a 3d cross-section of a 4d hypersphere and suggest in terms of set

theory that since no proton is unique that number of protons is actually number. This describes

the elements of the periodic table as sets of protons in a mathematical sense, and we suggest this

is geared around six-fold symmetry in a way that optimizes the conditions for biological life. The

radius of a proton is the least precise data value among the constants but new methods of

measuring it are honing in on its actual value. If our set theory for the elements in terms of

protons optimizes for life, then we have from our theory a value that is more within the current

finest measurements. We also suggest the unit of a second is a natural unit as luck would have it,

and that it occurs in the orbital elements of the Earth/Moon/Sun system. We determine a

constant k and intermediate mass that bridges the macrocosmos to the microcosmos that

predicts many things both on the atomic and astronomical and cosmological scales. It appears

our moon orbiting the earth may have connections on the atomic and cosmological scales. We

present the Solar System as quantum mechanical. In the second edition we come up with

another equation for the radius of a proton independent of Avogadro’s number which gives us

more equations for the unit of a second as natural to relate to the orbit of our Moon."

of 6 79

Introduction

The old enigma, the Universe is here, but it needs a cause. If we present a cause, then when

need a cause for that cause. Essentially it is as if the Universe is magic because it should not

be here; it is an uncaused cause. But, can we see that that can happen in terms of

mathematics?!

I had a theory for what matter is, or inertia. Mass is that which ﬁlls space over time. What is

space and time? We don’t know, it is our starting point and is not described by anything. For

instance what is velocity? It is the change in distance with time, meters per second (m/s). Let

us write velocity:!

Equation 1. !

But what is the distance x? It is velocity over time, that is how we deﬁne it. Let’s write that:!

Equation 2. !

Let’s put our value of velocity into equation 2:!

We have t cancels with t leaving x=x meaning distance is distance, so we have not said

anything about distance (spacial extension in one of the dimensions). We can in fact put all of

the relationships between mass, length, and time in any equation in physics and it will return!

So in the end we may have formally written out for instance an equation for gravity that will

land us on the Moon, but the equation reduces to saying nothing about what reality actually is. !

Back to my theory that is my theory for inertia. That theory is another subject, but I would like

to show how it suggests we can have something from nothing, which makes me wonder if this

mathematical trick I have found, is connected to how we might actually have something from

nothing, or that perhaps even something is nothing. Let’s lay the groundwork of that theory

brieﬂy:!

1.0. Ground Work

We suggest the property of matter to have inertia, which is to say it resists change that the

particle a proton, a fundamental unit of matter, is a the cross-section of a 4D hypersphere in 3D

space. As such one could consider inertia as the “friction” of space given by the normal vector

holding the space-bubble in a lower dimensional “plane”:!

Thus the mass of a proton is given by the force of space measured by G, the energy given

to it given by h, the speed at which things happen c, and the surface area of this sphere

v = x /t

x = vt

x =

m a ss = m a ss

distance = distance

time = time

4πr

of 7 79

that is the cross-section of the hypersphere and the ﬁne structure constant squared

because the fine structure constant squared is the ratio of the potential energy of an electron in

the first circular orbit to the energy given by the mass of an electron in the Bohr model times the

speed of light squared, that is it represents the ground state. It is

Equation 1.1

Thus the equation of inertia can be expressed as proton-seconds. We find that it is six-fold in

nature given the smallest integer for time (1-second) gives carbon the core element of life

chemistry, and 6-seconds is the largest integer before you get fractional protons. That is

Equation 1.2.

Equation 1.3.

From which instead of saying the left sides of these equations are seconds, we say they are

proton-seconds by not letting cancel with the bodies of these equations on the left, but

rather divide into them, which are in units of mass, giving us a number of protons. I say this is

the biological because as we shall see our equations are based on one second is 6 protons is

carbon, and 6 seconds is one proton is hydrogen, these making the hydrocarbons which are

the skeletons of biological life. We see this is a mystery of six-fold symmetry based around

biological life in the following computer program I wrote and its output:!

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydrogen(H )

h4π r

of 8 79

Is proton-seconds. Divide by time we have a number of protons because it is a mass divided by

the mass of a proton. But these masses can be considered to cancel and leave seconds. We

make a program that looks for close to whole number solutions so we can create a table of

values for problem solving.

By what value would you like to increment?: 0.25

How many values would you like to calculate for t in equation 1 (no more than 100?): 100

24.1199 protons 0.250000 seconds 0.119904 decpart

12.0600 protons 0.500000 seconds 0.059952 decpart

8.0400 protons 0.750000 seconds 0.039968 decpart

6.0300 protons 1.000000 seconds 0.029976 decpart

4.0200 protons 1.500000 seconds 0.019984 decpart

3.0150 protons 2.000000 seconds 0.014988 decpart

2.1927 protons 2.750000 seconds 0.192718 decpart

2.0100 protons 3.000000 seconds 0.009992 decpart

1.2060 protons 5.000000 seconds 0.205995 decpart

1.1486 protons 5.250000 seconds 0.148567 decpart

1.0964 protons 5.500000 seconds 0.096359 decpart

1.0487 protons 5.750000 seconds 0.048691 decpart

1.0050 protons 6.000000 seconds 0.004996 decpart

0.2487 protons 24.250000 seconds 0.248659 decpart

0.2461 protons 24.500000 seconds 0.246121 decpart

A very interesting thing here is looking at the values generated by the program, the smallest

integer value 1 second produces 6 protons (carbon) and the largest integer value 6 seconds

produces one proton (hydrogen). Beyond six seconds you have fractional protons, and the rest

of the elements heavier than carbon are formed by fractional seconds. These are the

hydrocarbons the backbones of biological chemistry. Here is the code for the program:

of 9 79

#include <stdio.h>

#include <math.h>

int main(int argc, const char * argv[]) {

int n;

ﬂoat value=0, increment,t=0, p=1.67262E-27, h=6.62607E-34,G=6.67408E-11,

c=299792459,protons[100],r=0.833E-15;

{

printf("By what value would you like to increment?: ");

scanf("%f", &increment);

printf("How many values would you like to calculate for t in equation 1 (no more than 100?): ");

scanf("%i", &n);

}

while (n>=101);

{

for (int i=0; i<n;i++)

{

protons[i]=((137*137)/(t*p))*sqrt(h*4*(3.14159)*(r*r)/(G*c));

int intpart=(int)protons[i];

ﬂoat decpart=protons[i]-intpart;

t=t+increment;

if (decpart<0.25)

{ printf("%.4f protons %f seconds %f decpart \n", protons[i], t-increment, decpart);

}}}}

We ﬁnd exactly our equation predicts the second as!

Equation 1.4 !

That this equals so perfectly one second leads us to suggest the second is a Natural Unit. We

ﬁnd it is, that it is in the kinetic energy of the Moon to that of the Earth times the Earth Day

which is the Earth rotation period:!

Equation 1.5. !

But that was using the average orbital velocities. We ﬁnd using aphelions and perihelions:!

Equation 1.6.

6α

h4π r

= 1.004996352secon d s

moon

Earth

(EarthDay) = 1.2secon d s

K E

moon

K E

earth

(Ear th Day) = 1.08second s

of 10 79

2.0. The Elements As Proton Sets

To get the Universe from nothing we consider equation 1.2 and 1.4:!

Equation 1.2. !

Equation 1.4. !

We can consider equation 1.2 to be two equations. The ﬁrst is where the numerator!

Cancels with the denominator!

Leaving:!

Equation 2.1. !

And the second equation is where does not cancel with the numerator but divides into it

giving us a number of protons times one second:!

Equation 2.2. !

Now we ﬁnd if we divide the left hand side by one second ( ) then the right hand side becomes

six protons.!

Equation 2.3. !

Now consider equation 1.4!

h4π r

= 6proton ⋅ secon d s = carbon(C )

6α

h4π r

= 1.004996352secon d s

h4π r

= k ilogra m s ⋅ secon d s

h4π r

= 6secon d s

h4π r

= 6proton − secon d s

h4π r

= 6proton s

of 11 79

It is approximately . Let us put it into equation 2.3:!

Equation 2.4. !

However factor 1!

Cancels with factor 2!

When this happens it says 6 equals 6 protons:!

The connection between number and form and they are the carbon atom that is at the core of

life. We can do the same thing with hydrogen (Equation 1.3):!

We see carbon is given by one second and hydrogen is given by six seconds ( equation 1.2)

Thus they are out of phase with one another by a factor of 6. We see this six-fold symmetry

produced hydrogen and carbon from nothing, and that they may in fact exist, but be nothing,

that they make up the skeletons of life that are the hydrocarbons, used as the basic chemical

structure underlying our mind that is consciousness. Thus we are conscious, but don’t really

exist. We seem to be conscious in a universe where the atom, the proton, is described by one

6α

h4π r

= 1.004996352secon d s

= 1secon d

h4π r

6proton ⋅ secon d s

1secon ds

= 6proton s

h4π r

6 = 6protons

h 4π r

= 1proton ⋅ 6secon d s = hydr ogen(H )

h 4π r

= 6proton ⋅ secon d s = carbon(C )

of 12 79

second, which came historically from our way of dividing up the calendar in ancient times

without knowledge of the atom in terms of the these modern constants. We also see this second

exists not just with the atom and its proton but with the Earth/Moon/Sun system which I found

was equations 1.5 and 1.6:

Equation 1.5. !

Using the average orbital velocities. We ﬁnd using aphelions and perihelions:!

Equation 1.6.

Earth day=(24)(60)(60)=86,400 seconds.

Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion we have:

Using the Moon’s orbital velocity at aphelion, and Earth’s orbital velocity at perihelion. We can

even use not the solar day used here (rotation of Earth with respect to the Sun) but the sidereal

day (rotation of Earth with respect to the stars) and the equation has even slightly more accurate

results.

In my work A Contemporary Anthropological Cosmology I derived the unit of a second in

terms of the hydrogen atom, and created a constant k that enabled me to predict the radius of a

proton with our equation 1.2 and 1.3 within experimental errors. But for now the point of this

paper is to show that either something can come from nothing, and that something is nothing,

and that this nothing actually creates life chemistry, if not physical chemistry in general, that

gives us consciousness, and that this consciousness has a connection with the Moon that it sees

in the sky from its planet Earth.

Interestingly the Moon perfectly eclipses the Sun which means as seen from the Earth the Moon

appears to be the same size as the Sun. This is because

Where is the Earth orbital distance, is the Moon’s orbital distance, is the Sun’s radius,

and is the Moon’s radius.

The Moon at its inclination to the Earth in its orbit makes life possible here because it holds the

Earth at its tilt to its orbit around the Sun allowing for the seasons so the Earth doesn’t get to

extremely hot or too extremely cold. We see the Moon may be there for as much of reason as is

Alpha Centauri, for a first manned mission beyond Earth.

moon

Earth

(Ear th Day) = 1.2secon d s

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

K E

moon

(7.347673E 22kg)(966m /s)

= 3.428E 28J

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

≈

⊙

of 13 79

If every proton in the universe is the same, none having any distinguishable features, that is

none are unique because they all have the same mass, same size, and same charge, then six

protons is the number 6 is carbon, and 1 proton is the number 1 is hydrogen, and carbon being

the number 6 tells us what the number 6 is, and hydrogen being the number 1 tells us what the

number 1 is. Because quanta such as protons and electrons are so small that the Heisenberg

uncertainty principle applies, then our proton, for instance, has no position in space only a

probability of existing in a given region at a given time. If then the proton has no position, it

cannot be distinguished from another proton by where it is in space. Thus I suggest the chemical

elements are sets, where each proton is the element of a set. Carbon is the set of six protons, and

hydrogen is the set of one proton. Because sets by set theory cannot have repeating elements, we

cannot say the set of protons that is carbon is

But rather we have to say

And the set of protons that is hydrogen is

We need not put the identifier 1 onto proton because hydrogen can never have more than one

proton because if it does it ceases to be hydrogen and is another element, like helium if you add

another proton. Hydrogen is thus the number 1 and does not need an identifier and 6 (carbon)

does need identifiers. I find this interesting because hydrogen cannot be identified as a metal or

non-metal but carbon can be identified as a non-metal. When a non-metal like carbon combines

with hydrogen it loses an electron to become thus ionizing like a metal and carbon gains 4

electrons to have noble gas electron configuration the same as neon (Ne) the noble gas in group

18 the last group of its period. But since carbon gains 4 electrons to become it must

combine with at least 4 hydrogens each to be neutral. This is which is methane gas.

This is C+4H=10 protons which is the element neon (Ne). This gives a time value of

Methane is the most basic hydrocarbon and hydrocarbons are the skeleton of biological life

chemistry. Methane gas was in the primordial Earth and when combined with ammonia ,

, and other gases thought to have been in the primordial Earth atmosphere, they make

some of the amino acids, the building blocks of life as was shown in the Miller-Urey experiment.

The radius of a proton is the least well known constant in our equation and is the subject of

much research to determine precisely. If our value of time t is supposed to be precisely

So as to center for it allowing for biological life then the radius of a proton is

C =

{

proton, pr oton, proton, pr oton, proton, pr oton

}

C =

{

proton1,pr oton2,proton3,pr oton 4,pr oton5,pr oton6

}

H =

{

proton

}

4−

t =

⋅

187769

1.67262E − 27

(6.62607E − 34)

6.67408E − 11

⋅

(0.833E − 15)

(299792459

= 0.602997811s

0.6 =

carbon

neon

of 14 79

This is in the range of the PRad (JLab) electron scattering method, a most recent method for

determining the radius of a proton, which is

3.0 Solving The Calendar Thus the duration of a second is!

Thus we developed a system to measure time from the motion of the Earth around the

Sun that creates a system where we measure time with a second that is 3 hundred

millionths of a year, the time it takes the Earth to go around the Sun once, and where

there are 86400 of these seconds (solar day) in the time it takes the Earth to rotate

once on its axis with respect to the Sun. With respect to the stars it takes 86164.0905

seconds. This is called the sidereal day. The solar day is a little longer because the

Earth has moved around the Sun a little before the Earth aligns with it again in its

rotation. It is this duration of a second that has given us!

Where 6 seconds is the largest integer before we get fractional atoms and 1 second is the

smallest integer that produces atoms and hydrogen and carbon are the hydrocarbons that make

the chemical skeletons of life chemistry.

That this equals so perfectly one second leads us to suggest the second is a Natural Unit. We

ﬁnd it is, that it is in the kinetic energy of the Moon to that of the Earth times the Earth Day

which is the Earth rotation period:!

But that was using the average orbital velocities. We ﬁnd using aphelions and perihelions:!

(6.67408E − 11)(299792459)

(6.62607E − 34)4π

18769

⋅ (1.67262E − 27 )(0.6) = 8.2885873E − 16m ≈ 0.829E − 15m

= 0.831E − 15m

2 %

1second =

365.25

⋅

= 0.0000000316881year

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydr ogen(H )

6α

h4π r

= 1.004996352secon d s

moon

Earth

(Ear th Day) = 1.2secon d s

of 15 79

These equations derive their meaning from classical and modern physics, yet the second came to

us from Ancient history of lost civilizations. We wish to present that as a mystery to add on to

these discoveries about Nature. We begin

It takes the Moon 27.3 days to go around the Earth, but since the Earth has orbited the Sun

since then it takes 29.5 days with respect to the Sun, or from New Moon to New Moon in other

words. Thus we say the lunar month is 30 days, the average of the 29 day month and the 31 day

month. I think I have solved the Calendar and why it is structured the way it is, why there are

365.25 days in a year. Well we know why that is, the earth rotates 365.25 times after it has gone

around the Sun once. But I think when making the calendar from which we got the second we

considered that 365.25 days is approximately 360 days, because we divide the circle into 360

degrees.

At the dawn of civilization, the Sumerians who settled down from wandering, gathering, and

hunting, to invent agriculture and and build ceramic homes, developed the first mathematics,

and it was sexagesimal (base 60) that passed on to the Babylonians, then ended up with the

Ancient Greeks, who divided not just the hour into 60 minutes of time and the minute into 60

seconds of time, but who divided the sky into hours, minutes, and seconds of arc which was

measured in time counted by the rotation of the Earth, and its orbital period around the Sun,

and the orbital period of the Moon around the Earth.

We are suggesting nature is founded on six-fold symmetry, six which is constructed by

multiplying together the first two prime numbers, 2 and 3. Two factorial is two, three factorial is

6. Two times six is twelve, the number of months in a year, approximated by our moon’s

approximately 12 orbits around the earth in the time it takes the earth to go around the Sun

once. There are four weeks in a lunar month, and

Is the number of seconds in the 24 hour Earth day. The duration of one second comes from

dividing up time and the sky like this in base 60, that came from the Sumerians, Babylonians,

and Ancient Greeks and it is believed they did this because 60 is evenly divisible by

1,2,3,4,5,6,…10,12,15,20,30,60,…

I think I have solved the calendar and why it is structured the way it is, and that it is not just

sexagesimal where 60 seconds and 60 minutes are concerned, but all the way through. We have

found the second pivotal to the atom and the Earth/Moon/Sun System in terms of sciences that

came long after the calendar’s formation. But, if we want to suggest the calendar was influenced

by advanced life from another planet, or that the Ancients lucked out because there was an idea

behind it, and we want to connect it to our equations of the proton and of the Moon and Earth

kinetic energies, which are dynamic, we have to find out why it is dynamic. I begin with the

second in terms of the calendar:

We write:

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 60

= 86,400

1secon d =

365.25

⋅

= 0.0000000316881year

of 16 79

There are actually 31557600 seconds per year, but we approximated 365.25 days with 360 days.

This however has an accuracy of

I then factor this into other ratios but keep the units:

Earth rotates through:

Moon moves around Earth:

Earth moves around Sun:

Thus our equation becomes

We then notice in one hour the earth rotates through 15 degrees giving

It does the same in 60 minutes. We also have

Seconds in a day: , and we write

(

360d a ys

year

)(

24h ours

d a y

)(

60min

h our

)(

60sec

min

)

= 31104000secon d s /year

31104000

31557600

= 98.56

(

60d a ys

year

)(

1h our

d a y

)(

60min

h our

)(

60sec

min

)

1h our = 60min

60d a ys =

year

360

∘

h our

360

∘

30d a ys

∘

d a y

360

∘

360d a ys

∘

d a y

60d a ys = 2(30d ays) = Lu n arOrbit

(

12d eg

d a y

⋅

d a y

deg

)

(

60d a ys

year

)(

1h our

d a y

)(

60min

h our

)(

60sec

min

)

24h ours

360deg

⋅ 15deg = 1h our

2(60

∘

) = 120

∘

2(120

∘

) = 240

∘

4(60

∘

) = 240

∘

(24h ours)(60min)(60sec) = 86400

sec

d a y

of 17 79

We have

86400sec

⋅ 360

∘

240

degrees

(

12d eg

d a y

⋅

d a y

deg

)

(

60d a ys

year

)(

24h ours

360

∘

⋅ 15

∘

)(

60min

h our

)(

60sec

min

)

(

60min

h our

⋅

60sec

min

)

3600sec

h our

3600

⋅ 360 =

degrees

(

24h ours ⋅

15deg

360deg

)(

60min

h our

)(

60sec

min

)

= 86400sec ⋅

∘

360

∘

15/360 = 0.013889

1/x = 72

86400

= 1200secon d s

of 18 79

Thus we have reduced the second to special triangles. The same second in our equation for the

proton, and for the kinetic energies of the Earth and Moon, and the rotation of the Earth. If we

can explain all three of these; the proton, the kinetic energies, and structural evolution of the

calendar, in terms of these triangles, we can speak of how they are all part of an underling idea

common to all three.

We have the following expression for seconds in a year:

We write out the angular velocities of the Moon and Earth in their orbits:

Put in the orbital periods of the Moon and the Earth and :

(

12d eg

d a y

⋅

d a y

deg

)

(

60d a ys

year

)(

24h ours

360

∘

⋅ 15

∘

)(

60min

h our

)(

60sec

min

)

12d eg

d a y

deg

d a y

(

)

(

2 ⋅ 30d a ys

year

)(

24h ours

360

∘

⋅ 15

∘

)(

60min

h our

)(

60sec

min

)

(

)

(

)

(

⋅

∘

360

∘

)(

60min

h our

)(

60sec

min

)

(

)

(

)

(

⋅

∘

360

∘

)(

60min

h our

)(

60sec

min

)

(

)

(

)

(

∘

)(

1min

h our

)(

60sec

min

)

(

)

(

) (

)

(

60sec

h our

)

(

)

(

)

(

300sec

h our

)

1h our = 3600secon d s

of 19 79

Which is true because it is

And we had made the approximation that there are 360 days in year because there are 360

degrees in a circle. We see the calendar divides units of measurement into base 60

Which describes the motion of the moon and the Earth

In the literature it says we use base 10 probably because we have 10 fingers to count on. But 10 is

2 times 5 and 2 is the smallest prime number. Base 60 is 10 times six and sixfold symmetry is

the most dynamic because it is 2 times 3 and 2 and 3 are the smallest prime numbers one even,

one odd. The regular hexagon, a six-sided polygon has its radii equal to its sides because it

divides-up into equilateral triangles. But I suggest we have base 10 not just because we have 10

fingers to count on but because 2 times five is 10 and 10 times 6 is 60 and 60 is evenly divisible

by so much. Evenly divisible by

1,2,3.4.5.6….10,12,15,20,30,60

It is known that the physical like snowflakes are based on 6-fold symmetry and life is based on

5-fold symmetry like starfish or two arms, two legs, and a head. More often than not a flower

will have 5 petals. If it has hundreds like a rose, it will be based on the golden spiral and the

golden ratio is derived from the regular pentagon (five-sided polygon). Thus we have not just

solved the calendar, but found more reasons why we have base 10 counting. It alls has to do the

physical, that of six, and the organic, that of five.

(

)

(

)

(

300sec

3600sec

)

(

)

(

)

(

)

= 12

(

)

(

)

secon d s

year

(

)

(

30d a ys

)

= 360d a ys

(

12d eg

d a y

⋅

d a y

deg

)

(

60d a ys

year

)(

24h ours

360

∘

⋅ 15

∘

)(

60min

h our

)(

60sec

min

)

(

)

(

)

secon d s

year

of 20 79

Since we have

Then it follows

We want to write this in terms of the sizes of the Moon and the Sun, the masses of the Earth and

the Sun, and the orbital radii of the Moon and the Earth.

Essentially that the Moon perfectly eclipses the Sun as seen from the Earth means while it is 400

times smaller than the Sun it is 400 times further from the Sun than it is from the Earth. This

determines its orbital velocity and mass, as well as that of the Earth and the mass of the Sun.

The orbital velocities of the Moon and the Earth are given by:

and

The Moon perfectly eclipses the Sun because

Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius, and

is the lunar radius. This gives:

We have more explicitly:

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

(

)

(

)

secon d s

year

(

)

(

)

(

K E

)

(

Ear th Day

)

= 1secon d

⊙

⋅

⊙

⋅

1.989E 30kg

5.972E 24

⋅

1.74E6m

6.957E8m

= 28.86

≈

⊙

⋅

⊙

= 28.6

of 21 79

The rotational period of the moon is is the same as its

orbital period, and the rotation of the Earth is is the ratio

27.3 which gives

From

We have

We want to see how accurate this is:

But the radius of the Sun is difficult to pinpoint. Beyond the bulk of its mass it has a thin

atmosphere that reaches far out into space. However, the Moon nearly perfectly eclipses the Sun

as seen from the Earth allowing us to study its outer atmosphere during an eclipse. The moon

nearly perfectly eclipses the Sun as seen from the Earth because the Sun may be 400 times

larger than the Moon, but the Moon is 400 times further from the Sun than it is from the Earth.

This is to say that

Where is the Earth orbital radius, is the lunar orbital radius, is the solar radius, and

is the lunar radius. So why not let the size of the Moon as seen from the Earth define the radius

of the Sun by writing

⊙

⋅

⊙

⋅

1.74E6m

6.957E8

1.989E 30kg

5.972E 24kg

⋅

1.496E11m

3.84748E8m

= 28.46

= 27.3d ays = 2,358, 720secon d s

= 24hours = 86,400secon d s

⊙

⋅

(

)

(

)

(

K E

)

(

Ear th Day

)

= 1secon d

⋅

(

)

⊙

⋅

K E

(

Ear th Day

)

= 1secon d

31557600s

86400s

⋅

(

)

7.1544E8m

1.7371E6m

5.972E 24kg

1.989E 30kg

⋅

3.6323E8m

1.496E11m

(1.08s) = (1.07)(1.08s) = 1.1556 s

≈

⊙

of 22 79

The Moon at closest approach to Earth (perigee) is 1.7371E7 meters:

This is the radius of the Moon I used to verify our equation. But let us take

And verify it with

Equal to its exact value for the Earth year, and the sidereal month of the moon.

=0.9856deg/day

And our results just become better. Now we just have one more thing to do, to take our equation

for the second in terms of the proton

And!

To get,

1.496E11m

3.632289E8m

⋅ 1.7371E6m = 7.154446E8m

⋅

(

)

⊙

⋅

K E

(

Ear th Day

)

= 1secon d

⋅

(

)

365.25d a ys

360

∘

27.3d a ys

360

∘

= 13.1868deg /d a y

31557600s

86400s

⋅

(

)

7.1544E8m

1.7371E6m

5.972E 24kg

1.989E 30kg

⋅

3.6323E8m

1.496E11m

(1.08s) = (0.988)(1.08s) = 1.067s

6α

h4π r

= 1.004996352secon d s

⋅

(

)

⊙

⋅

K E

(

Ear th Day

)

= 1secon d

⋅

(

)

⊙

⋅

K E

(

Ear th Day

)

6α

⋅

h 4π

of 23 79

4.0 Predicting The Meter In the last section we took our equation for a second as given by

the calendar!

And found we could write it in sexagesimal (base 60)

Which resulted in

Which we found was

Because

We then used that the Moon perfectly eclipses the Sun

And Universal gravity for the Earth and Moon

and.

To get

And since we had for the proton!

1secon d =

365.25

⋅

= 0.0000000316881year

(

60d a ys

year

)(

1h our

d a y

)(

60min

h our

)(

60sec

min

)

(

)

(

2 ⋅ 30d a ys

year

)(

24h ours

360

∘

⋅ 15

∘

)(

60min

h our

)(

60sec

min

)

(

)

(

)

secon d s

year

(

)

(

30d a ys

)

= 360d a ys

≈

⊙

⋅

(

)

⊙

⋅

K E

(

Ear th Day

)

= 1secon d

6α

h4π r

= 1.004996352secon d s

of 24 79

We ended-up with!

We now want to put certain quantities in parenthesis:

We have the key orbital elements and the equation of a proton:

We want to consider

411.858 refers to the Sun being 400 times larger than the Moon allowing for an eclipse because

the Moon is 400 times further from the Sun than from the Earth. Since we have rounded the

year to 360 days because there are 360 degrees in a circle and we have rounded the lunar

angular orbital velocity to 12 degrees and that of the Earth to 1 degree, both per day, then let us

write

Then 0.000085383 becomes x=0.000083333 and we have

We notice that the radius of a proton is . Thus we have

⋅

(

)

⊙

⋅

K E

(

Ear th Day

)

6α

⋅

h 4π

(

)(

)

⊙

⋅

(

K E

(Ear th Day)

)

6α

⋅

h 4π

(

)

360d a ys

1d a y

(

K E

(Ear th Day)

)

= 1secon d

(

)

∘

/d ay

∘

/d ay

6α

⋅

h 4π

= 1secon d

⊙

⋅

7.15448m

1.7371E6m

5.972E 24kg

1.989E 30kg

⋅

3.6323E8m

1.496E11m

= (411.8588)(0.000085382)

(

360

)

(

)

(

400

) (

1sec

)

= 1sec

⊙

⋅

= 0.000083333

= 0.833E − 15m

0.000083333

= 1.000E11m

−1

of 25 79

We know the average orbital distance of the Earth from the Sun is an astronomical unit (AU)

which is 1 AU=1.496E11m. That is because the orbit is close to circular. Thus we have

Using the orbital velocity of the Earth at perihelion (closest approach to the Sun when it is going

fastest) and the average orbital velocity of the Moon, , we have

As you can see

And you will find there is little difference of you use the sidereal lunar month which is 86,164.1

seconds. You get 1.20687 seconds, which is even closer, in fact almost exact.

We call upon

And have!

= 1AU

⊙

⋅

1.989E 30kg

5.972E 24kg

⋅

1.496E11m

3.6323E8m

⋅ (0.833E − 15m)(1.496E11m) = 1.4595m

1.45954m

= 1.208m ≈ 1.2m

= 30,290m /s

= 1,022m /s

K E

(Ear th Day) = 1.2s

K E

(7.3476731kg)(1,022m /s)

= 3.83726E 28J

K E

(5.972E 24kg)(30,290)

= 2.7396E 33J

3.83726E 28

2.7396E 33

(86,400) = 1.210s

(

⊙

⋅

⋅ r

)

(Ear th Day)

1.2m eters

1.2secon d s

= 1.0

6α

h4π r

= 1.004996352secon d s

(

⊙

⋅

⋅ r

)

(Ear th Day)

⋅

6α

h 4π

= 1.0m eters

of 26 79

5.0 The Constant k We introduce a constant k, which is determined by the velocity of the

Earth:!

5.1 !

Six for six-fold symmetry. To derive this we introduce Giordano’s relationship. !

5.2 !

The number on the left is the number on the right, but with diﬀerent units. I found I could

eliminate the and gain six-fold symmetry by creating an equation of state for the periodic

table of the elements using Avogadro’s Number ( ). We say that!

5.3

And,

5.4

Which is basically true, hydrogen is one proton and the mass of the electron is nominal. For

every 6E23 protons of hydrogen, there is one gram. Then we always have

5.5

We can say for any element

Where is the number of protons in the element, so for carbon

Thus by equation 5.2 we have

5.6

k v

= 6

h(1 + α) ⋅ 10

= G

6.02E 23

= 6E 23

proton s

gr a m

ℍ = 1

gr a m

proton

⋅ 𝔼 = 6E 23

𝔼

Z ⋅ 6E 23proton s

Z ⋅ gr a m s

𝔼 =

Z ⋅ gr a m s

Z ⋅ proton s

ℂ

ℂ =

6gr a m s

6proton s

6(6E 23proton s)

6gr a m s

ℂ = 6E 23

(1 + α)

⋅ N

𝔼 = 6.00kg

⋅

of 27 79

And we have introduced our 6 of six-fold symmetry. We need an intermediary mass between the

microcosmos the proton mass and the mass of a star M. We use as the limiting factor for the

mass of a star as the Chandrasekhar limit, the upper limit of mass for a white dwarf star to form

without collapsing into a blackhole star. It balances with its gravity by radiation pressure alone.

It is:

5.7

We say the intermediary mass, , is

5.8

Inserting 5.7 into 5.8, we have

5.9

Where we made the approximation . This results in

5.10

We can hone this by reintroducing 0.77 for 3/4

Thus precisely:

We have

We have honing our

Multiplying 6 by we have our value for k.

M ≤ 0.77

ℏ

= 1.41 ⊙

= Mm

ℏ

1/2

0.77 ≈ 3/4

8π

1/2

= 67.9943kg

= 0.8660

0.77 = 0.8775

0.8775

0.8660

= 1.01328

= (67.9943)(1.01328) = 68.897kg ≈ 69kg ≈ 70kg

1/m

of 28 79

5.11

This gives a value of k as

5.12

5.13

Putting this in equation 5.1 we have

5.14

Which is very close to our prediction of six-fold symmetry. is the average orbital

velocity of the Earth. I find this constant k predicts the duration of the Universe in the

standard Friedman Model. We can write

5.15

This is

5.16

The constants suggest a basic Universal Energy by dimensional analysis

5.17

So we can write

5.18

1 Earth year = (365.25)(24)(60)(60)=31557600 seconds

k =

(1 + α)

⋅ N

𝔼 =

773.5

⋅

k =

(68.897kg)

(6.62607E − 34) ⋅

(1.007299)

6.67408E − 11

⋅ 6.02E 23 = 0.001268291s /m

= 788.4626

k v

29790m /s

788.46m /s

= 6.145748

KE = m

(

)

KE = (68.897kg)(788.4626m /s)

= 42,831,358Joules

⋅

= 1.599298E 29J

6.62607E − 34

6.67408E − 11

⋅

299792459

1.67262E − 27

= 1.599298E 29J

⋅

(

)

(1secon d ) =

1.599298E 29J

42831358J

= 3.734E 21secon d s

of 29 79

The universe is theorized by standard models to die in 100 trillion years, which is when the last

stars born will die out. This is exactly 1E14 years. We have

5.19

We ﬁnd as well that k predicts the Earth year. We suggest for some mass , we have

It is given by

5.20

For all practical purposes this is the mass of the Moon, which is exactly 7.34767E22kg.

We have:

3.734E 21s

31557600s

= 1.1832332E14years ≈ 1E14years

⋅

(

)

(1secon d ) = Li feSpa nUniverse

K E

Earth

K E

moon

(1secon d ) = E ar th Da y

6α

h4π r

= 1.004996352secon d s

⋅

(

)

(1secon d ) = 1E ar thYear

M =

⋅

(1Ear thYear)

(

)

(1secon d )

⋅

6.62607E − 27

6.67408E − 11

⋅

(299792459)

1.67262E − 27

= 1.599E 36J ≈ 1.6E 36J

= (788.4626)

= 621,673.27

1Ear thYear = (365.25)(24)(60)(60) = 31557600s

M =

1.599E 36

621673

⋅

31557600

= 8.15E 22kg

8.15E 22kg

7/34767E 22kg

= 1.10919516 ≈ 1.12m oon s

of 30 79

5.21

We also ﬁnd that k predicts the radius of a proton. We have the radius of a proton is given by

carbon by evaluating at one second:

5.22

But to get that we have to multiply by one second and we need one second in terms of the atom

for a theory of the proton. I find we can do that…

5.23

5.24

Substitute for to get

5.25

Where is the Van Der Waals radius for a hydrogen atom. We have now introduced the

radius of a hydrogen atom . Our formulation of inertia as proton seconds is a

form of impulse. To change that to momentum we have to divide by a second. This radius of the

hydrogen atom is the Van Der Waals radius, which is the closest distance between two hydrogen

atoms non-covalently bound. It is 120 pm. Divide that by where 1/k is our constant

⋅

(

)

(1secon d ) = 1Ear thYear

Earth

moon

(1secon d ) = Ear th Day

6α

h4πr

= 1.004996352seconds

4πh

= 8.288587 × 10

−16

m = 0.829f m

6α

⋅

h 4π r

6α

⋅

h 4π r

h 4π

t =

2α

h 4π

= 1.2E − 10m

⋅

8π

⋅

1 + α

⋅

𝔼

≈

⋅

8π

⋅

𝔼

of 31 79

And we find

5.26

5.27

We have our equation for the radius of a proton

We only need to multiply it by to have the right units, and we get

5.28

Then suggest we picked up 9/8 in approximations which is close to one anyway so we write

5.29

We form constants:

And we have the Equation:

5.30

We can say that Avogadro’s number is not an arbitrary number because it is such that there are

twelve grams of carbon and carbon is 6 neutrons plus 6 protons equals 12. We now want to add

to this section the equation of the mass of a proton.

3 2

⋅

6.626E − 34

6.674E − 11

⋅

18769

(1.6726E − 27)

⋅

1.2E − 10

6.02E 23

⋅

= 1.12secon d s

3 2

⋅

Gπ

⋅

𝔼

= 1.12secon d s

4πh

= 8.288587 × 10

−16

m = 0.829f m

t /ck = 1secon d

2 ⋅

4π

⋅

𝔼

2π

⋅

𝔼

= 8.26935E − 16m ≈ 0.827f m

(

)

(

)

(

)

ℍ

(

)

k =

2π

= 6.93E − 9kg

𝔼

= 1.99E − 34m

= k

𝔼

of 32 79

5.31

If our equation is right and we put it into natural units then the product should be close to

one:

Let us start with the units with which we are working:

And convert these to proton-masses and proton-radii:

Now we find k in these units:

Thus we have from equation 30:

2π

⋅

𝔼

G =

kg ⋅ s

h = kg ⋅

c = m /s

G = 6.67408E − 11

kg ⋅ s

⋅ 1.67262E − 27kg(0.833E − 15m)

⋅ s = 193,131, 756

h = 6.62607E − 34kg ⋅

⋅

(0.833E − 15)

(1.67262E − 27kg)

= 5.71E 23

c =

(299,792, 459m /s)(1sec)

(0.833E − 15m)

= 3.6E 23

1.2E − 10m

0.833E − 15m

= 144,058

k =

2π

= 6.93E − 9kg

k =

(5.71E 23)(3.6E 23)

2π

(193131756)

= 4E18proton − m a sses

= k

𝔼

(4E18)(144058)

(6E 23)

5.76E 23

(6E 23)

= 0.96 ≈ 1

of 33 79

We can also predict the charge of an electron with k…

I construct the electric field as such (Fig. 2): is one component of . Events in

are through time with components and where c is the speed of light. The

electrons and the protons and in the presence of one another cross into giving

themselves acceleration in the and directions.

We have

We suggest

5.32

The position of changes in the space of and has travelled to a place in

time, as well. Since ct=meters there is a g such that . Thus since

ℝ

(x, y, z)

ℝ

(x, y, z)

−x



0 ct

0 0 ct

= (ct

− ct

)



··

x ∝ (ct

− ct

)



ℝ

(x, y, z)

t (0,ct

, ct

)

··

x = g(ct

− ct

)



g ⋅ m =

Fig. 2

of 34 79

and that is . Thus g is frequency squared ( ) and we suggest it is

derived from the separation between the charges and . That there is some velocity v such

that . We call upon our equation for 1 second:

Letting v= from our equation for k

And we have

5.33

We get

5.34 !

g =

⋅

⋅ s =

f =

(

)

h 4π

= 1secon d

(

)

(c)

h 4π

⋅

(c)

h 4π

= k

⋅

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

= 773.5

m = m

(

⋅

)

h 4π r

= q

h 4π r

(6.626E − 34)(4π)(0.833E − 15)

(6.674E − 11)(299,792, 459)

= 5.37E − 31

q = (0.033)

36(18769

)

773.5

(5.37E − 31) = 9.1435E − 19C

= ⋅

9.1435E − 19C

1.602E − 19C

= 5.71proton s ≈ 6protons

of 35 79

6.0 Discussion

When factor 1 and factor 2 cancels (page 11) it says 6 equals 6 protons. The connection between

number and form.

Interestingly this harkens back to the Logical Positivists. At the time mathematicians realized

nothing had any meaning that we didn't even know what 1, or 2, or 3, or 4....meant. So Bertrand

Russel set out to make a mathematics that explained it. It was said 4 was 4 objects and and set

theory was invented to show that a set was a collection of objects, for which its inventor Georg

Cantor made rules to explain what an object was. But in the end the British mathematician

Bertrand Russel showed it to contradict itself, then Gödel came along and proved that all formal

systems have at least one unprovable statement, and are thereby incomplete with his

Incompleteness Theorem. So they gave up. I think this is connected to 6 equals 6 protons in my

theory. They couldn’t say what an object was though it was tried to establish that it could be

anything but here it may be a proton and I think that works because any object is made of

protons, and protons are the basic units of matter and are uniformly the same. I am getting

interested now in this set theory and Gödel's theorem. But the recently added chapter on solving

the calendar has got me very intrigued as well, just where the Ancients got it, or if it evolved to

something naturally on its own.

The logical positivists may have given up on logical positivism, but set theory is used today in

applications even though it never said what an object is or a number. I think it is interesting that

I get 6 is 6 protons, is carbon or hydrogen is 1 is 1 proton because objects are made up of

protons. Interesting are the quarks which are inferred from experiment, but cannot owning to

the principle of color confinement, exist in isolation, they must be found in hadrons. Hadrons

are protons and neutrons. But the proton can exist in isolation.

of 36 79

7.0 Solar System A Quantum Mechanical System The constant k in this paper is a sort of

intermediate velocity in the Universe, and so is the intermediary mass used to determine it.

While there are small masses in the universe like pebble sized rocks and debris in the asteroid

belt there are larger masses like planets. The intermediary mass is determined by a proton and

white dwarf star. It combined with the intermediary velocity 1/k which gives an energy that

predicts the life span of the universe in the standard Friedman model predicted by when the

last stars would be made and burn out from the mass of the Universe.!

This intermediary mass and energy seems to be associated with the masses and speeds in the

human realm between atoms and stars, like a mean for mass and velocity in the universe. It

predicts the radius and charge of a proton as well as the life span of the universe. But I asked

what do we know of that has exactly these masses and velocities. !

The intermediary mass is 68.897 kg is about 152 pounds. I knew a wolf is less than this a

jaguar more, but I found it is exactly an adult male mountain lion, a cougar.!

But what speed is 1/k=788.4626m/s. It is 1,764 miles per hour. That is the speed of a fast

military jet. it is about mach 2 or twice the speed of sound, which is about 767 miles per hour.

Our fastest jet (record holder) went mach 6.7. But it was a real milestone to achieve mach 2

which was ﬁrst achieved by the Lockheed F-104 Starﬁghter.!

The fastest running animal is the Cheetah. So it would be like the record holding jet at mach

6.7. But the Cougar for which the adult male is nearly exactly the intermediary mass in the

Universe, would be like the F-104 Starﬁghter which goes almost exactly 1/k.!

We have suggested that the unit of a second and the unit of a meter are Natural and pertain to

the structure of the Universe. But what would be the Natural unit for mass. We could suggest it

is a kilogram, but that does not necessarily make sense. We have described the second in

terms of the proton!

And in terms of the Moon/Earth/Sun system!

And the meter in terms of the proton Earth/Moon/Sun system!

We want to describe the kilogram as a Natural unit. We already have the kilogram described as a

natural unit. It is described in terms of the density of water at standard temperature and

pressure:

6α

h4π r

= 1.004996352secon d s

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

(

⊙

⋅

⋅ r

)

(Ear th Day)

⋅

6α

h 4π

= 1.0m eters

of 37 79

But we want something more natural (Sure water is natural and a primary ingredient of life, and

standard temperature and pressure is ideal for life) but we want things in terms of the proton

and Earth/Moon/Sun system because we want to suggest the Earth/Moon/Sun system has a

Universal idea behind it and the proton is already Universal.

The mass of the Moon

The mass of the Earth

The intermediary mass

This needs to be multiplied by 1.2 to get 1 kg. But remember

It is actually closer to 1.25 seconds. And, remember

Thus our equation is

, ,

Using mean orbital velocities for the Earth and the Moon

1kg ≡ ρ

H2O

⋅ m eter

= 7.34767E 22kg

= 5.97219E 24kg

= 68.897kg

) = (0.0123)(68.897kg) = 0.847kg

K E

(Ear th Day) = 1.2secon d s

6α

⋅

h 4π

= 1.004996352secon d s

6α

⋅

)

K E

(Ear th Day) ⋅

h 4π

= 1kg

α = 1/137

1/α

= 1/18769

6/α

= 0.00032

K E

(5.97219E 24kg)(29,790m /s)

= 2.65E 33J

K E

(7.34767E 22kg)(2,033m /s)

= 3.837E 28J

K E

(Ear th Day) =

3.837E 28

2.65E 33

(86,400s) = 1.251E 28secon d s

of 38 79

Remember using the sidereal day just makes it closer to 1.2s but makes little difference in the

results. We have

Thus for units of mass, length, and time as Natural Units we have…

And we point out that (because it really is quite incredible) for the unit of time the second is

described by just the proton, independently of the Earth/Moon/System and is described by just

the Earth/Moon/System, independently of the proton. In the case of the meter and the kilogram

they are described in terms of both, not independently of one another. The incredible thing

about the second is it comes from the evolution of the calendar since ancient times by

reconciling the periodicities of the Moon and Sun in the Earth sky by using sexagesimal, or base

60 counting, which we already went into in full. We have that

Then force, F, or is

(0.847)(1.251)

1.005s

= 1.054kg ≈ 1.0kg

6α

h4π r

= 1.004996352secon d s

K E

(Ear th Day) = 1secon d

(

⊙

⋅

⋅ r

)

(Ear th Day)

⋅

6α

h 4π

= 1.0m

6α

⋅

)

K E

(Ear th Day) ⋅

h 4π

= 1kg

(

⊙

⋅

⋅ r

)

(Ear th Day)

= 1.0m /s

6α

(

⊙

⋅

⋅ r

)

(Ear th Day)

4πh

= 1.0m /s

= 1N

of 39 79

We finally have energy, Joules, or is

Something very interesting is going on here so we want to start a new page to look at it so all the

equations are on one page…

36α

⋅

)

⊙

⋅

⋅ r

1/2

4πh

= 1N

J = kg

6α

)(m

)

4πh

⋅

⊙

⋅ r

(Ear th Day)

= 1J

of 40 79

Let us look at our equation for kilograms. This is actually incredible. Let’s see that: We write it

And say that

The mass above is the intermediary mass adjusted by the ratio of the Moon’s mass to the Earth’s

mass. The spin is given by the rotation of the Earth adjusted by the kinetic energy of the Moon to

the kinetic energy of the Earth and is close to one second, and in some cases is one second. The

frequency is given by the size and mass of a proton and is one cycle per second, its inverse is

actually one second. The product of all three is actually one kilogram even. The second turned

out to be a natural unit, and now so is the kilogram. We want to point out that the intermediary

mass is, where we made the approximation 0.77~3/4:

The constant k is

Which putting in is

And finally we see k with the velocity of the Earth defines the sixfold symmetry in this

(

)

)(

K E

(Ear th Day)

)(

6α

⋅

h 4π

)

= 1kg

(

)

= m a ss = 0.847kg

(

K E

(Ear th Day)

)

= spin = 1.2secon d s

(

6α

⋅

h 4π

)

= f requen c y = 1secon d

−1

8π

1/2

= 67.9943kg

k =

(1 + α)

⋅ N

𝔼 =

773.5

⋅

k =

8π

(1 + α) ⋅ N

𝔼

k v

= 6

(

)

= (68.897kg)(788.46m /s)

= 4.92558E6J ≈ 5,000,000J

of 41 79

In so-much-as we have

Where

And

Is like a quantum number where

And

Since quantum mechanical states are described by mass, spin, and frequency, it may be that the

Earth/Moon/Sun system is a quantum mechanical system. Which is interesting because the

Earth is the planet in our solar system brimming with life and the Moon makes that possible by

holding the Earth at the inclination to its orbit around the Sun that it has allowing for the

seasons preventing extreme hot and extreme cold. Thus we want to introduce the energy

And note that in a quantum mechanical system, the energy of a quantum is

Where is Planck’s constant and is frequency.

(

)

)(

K E

(Ear th Day)

)(

6α

⋅

h 4π

)

= 1kg

(

)

= m a ss = 0.847kg

(

K E

(Ear th Day)

)

= spin = 1.2secon d s

(

6α

⋅

h 4π

)

= f requen c y = 1secon d

−1

k v

= 6

k =

(1 + α)

⋅ N

𝔼 =

773.5

⋅

8π

1/2

= 67.9943kg

E = m

(

)

= (68.897kg)(788.46m /s)

= 4.92558E6J ≈ 5,000,000J

E = h ν

of 42 79

We apply our formulation to predict the Earth orbital period as a quantum mechanical system.

In so far as we have for the solar system

And, for the the atom

We introduce what we will call a universal energy

The mass of the Moon is

We have

This is the Earth year (Earth orbital period) to an accuracy of

90%

E = m

(

)

= (68.897kg)(788.46m /s)

= 4.92558E6J ≈ 5,000,000J

E = h ν

⋅

6.62607E − 27

6.67408E − 11

⋅

(299792459)

1.67262E − 27

= 1.599E 36J ≈ 1.6E 36J

= 7.34767E 22kg

E = M

(

)

= 7.34767E 22kg(788.4626m /s)

= 4.56785E 28J

⋅

(

)

⋅

6α

⋅

4πh

1.599E 36J

4.56785E 28J

⋅ 1secon d = 35005527secon d s

(365.25d a ys)(24hrs /d a y)(60m /hr)(60sec /min) = 31557600secon d s

31557600

35005527

of 43 79

8.0 Solar Luminosity A QM State That I have found the Earth/Moon/Sun system is

quantum mechanical like the atom, is interesting because we ﬁnd the orbital period of the Earth

around the Sun from the mass of the Moon orbiting it, alone. In classical mechanics, to get the

orbital period of the Earth you have to use the mass of the Sun and the Earth’s distance from it.

To explain this curious phenomenon that I have found regarding the Earth I am looking at why it

is the distance it is from the Sun because that is given by the orbital period (Kepler’s law,

). I am suggesting it is the distance it is from the Sun because it allows for life (puts it

in the habitable zone, where water exists in its liquid phase abundantly at a good temperature).

This is determined by the luminosity of the Sun. The luminosity of the Sun at its surface is

determined by its temperature at its core which is determined by its mass and radius. But the

temperature at the surface comes from the temperature at the core cooling by the time it

reaches the surface. We want to explain!

That is, the Earth year is given by the Moon’s mass because The Earth is ideal for life. The

temperature of the Sun at its core is described by!

1. !

= !

But we don’t want to just consider the mass of hydrogen ( ) but the mass of helium as well.

The sun is 91.0% hydrogen and 8.9% helium by number of atoms. We have!

0.91(1.67E-27)=1.5197E-27!

0.089(1.67E-27)4=5.9452E-28!

1.519E-27+5.9452E-29=2.11422E-27~2.11433E-27kg!

Heat dissipates by an inverse square law. The core of the Sun is 1.33E6m and its radius is

6.957E8m. Thus the temperature at its surface would be if the only factor was the inverse

square law (which it isn’t, it is complex and requires the inclusion of many factors):!

2. !

= !

= a

⋅

(

)

⋅

6α

⋅

4πh

= 1year

⊙

⋅

⊙

1.67E − 27kg

6.957E8m

⋅

6.67E − 11

1.38E − 23

1.989E30kg = 2.3E 7

∘

2E − 27kg

6.957E8m

⋅

6.67E − 11

1.38E − 23

1.989E30kg = 2.7637E 7

∘

(

)

⊙

(

1.22E6m

6.957E8m

)

(2.7637E 7 ∘ K )

of 44 79

= !

The actual temperature at the surface of the Sun is 5,780 degrees K. So we are oﬀ by a factor

of!

It is at this point that we suggest we can bypass the complex factors that also aﬀect the

surface temperature of the Sun by using our Quantum Mechanical system for the solar system.

Not that the Sun isn’t described by Quantum mechanics in looking at its atoms, but in that our

Quantum Mechanical system treats the sun as if it is one proton with planets orbiting it. To do

this we notice that the factor of 68 is very close to our intermediary mass . But that is

in kilograms and our factor is dimensionless. But we developed a quantum mechanical

description of 1kg in the last section for the Earth/Moon/Sun system. It was!

3. !

Equation 1, 2, and 3 yield!

4. !

=(0.000003075)(2.7637E7deg K)((81.28)(0.804)=5553.6 deg K!

The luminosity of the Sun is given by!

5. !

= !

The actual luminosity of the Sun is 3.846E26 Watts. This was using the second given by Earth

and Moon kinetic energies as 1.2 seconds from average velocities. We can go as low as 1.08

seconds using apogees and perigees. And we can get values in between because though the

orbits of the Moon and Earth are close to circular, they aren’t exactly, so there is slight

variations in the result, so the above wavers around the exact answer.!

We want to consider how much sunlight reaches the Earth. We use our value for solar

luminosity:!

84.98975 ≈ 85

∘

5780

= 68

68.897kg

(

)

)(

(Ear th Day)

)(

6α

⋅

h4π

)

= 1kg

(

)

(

⊙

⋅

⊙

)

6α

⋅

4πh

⋅

(Ear th Day)

(

1.22E6m

6.957E8m

)

(

2E − 27kg

6.957E8m

⋅

6.67E − 11

1.38E − 23

(1.989E30kg)

)(

5.97219E 24kg

7.34767E 22kg

)

1.005s

1.25s

⊙

= 4πR

⊙

σ T

4π(6.957E8m)

(5.67E − 8)(5553.6

∘

K )

= 3.28E 26Wat ts

= 3.28E 26J/s

of 45 79

The separation between the Earth and the Sun is!

The solar luminosity is reduced at Earth by the inverse square law giving the solar constant:!

intercepts the Earth disc and distributes itself over the entire surface and

because the Earth’s albedo, , is 0.3 it reﬂects 30% of the light back into space. We have!

Temperature to the fourth is proportional to radiation by the Steﬀan-Boltzmann constant, :

means temperature entering:!

So we have the annual average temperature would be!

The solar constant is usually 1361 (Well established). Let’s convert this to centigrade:!

The actual annual average temperature of the Earth is 14C or 57F. But we have not included

the eﬀect of the Earth’s greenhouse gases holding in heat. In a one layer atmosphere model we

consider that the radiation entering the system equals the radiation leaving the system when

the Earth is in radiative equilibrium. The atmosphere radiates back to the surface. So

the temperature at the surface of the Earth is!

= 1.496E11m

3.28E 26

4π(1.496E11m)

= 1166

Wat ts

meter

πR

4πR

(1 − a)S

πR

4πR

= (1 − a)

σ T

(1 − a)

4σ

(1 − a)

σ = 5.67E − 8

1166

4(5.67E − 8)

(0.7) = 245K

= 245K − 273.15K = − 28.15

∘

(−28.15)(1.8) + 32 = − 18.67

∘

= T

σ T

= σ T

+ σ T

= 2σ T

of 46 79

So,…!

So this is hotter than the annual average temperature we said was 14C or 57F but we have not

considered cooling by convection; A lot of the radiation goes into warming the ocean, which

has a high speciﬁc heat (4.184J/g-K) and when this water evaporates it precipitates and returns

as rain. If we include this we get much closer to accurately predicting the global temperature.

There are other cooling mechanisms to consider as well. !

We have done what we set out to do to show it is possible that the Earth year is given by the

mass of the Moon because the Earth/Moon/Sun system is like a quantum mechanical one that

provides a state optimum for life. We have shown!

Where , hydrogen, is weighted with , helium according to their relative abundances by

number of atoms of each.

= 2

= (1.18921)245 = 291.356K

= 302.75 − 273.15 = 18.206

∘

18.206(1.8) + 32 = 64.771

∘

⋅

(

)

⋅

6α

⋅

4πh

= 1year

(

)

(

⊙

⋅

⊙

)

6α

⋅

4πh

⋅

(Ear th Day)

of 47 79

9.0 Genesis Project Equations We have for the Earth/Moon/Sun system for there to be life on

Earth that the temperature of the Sun at its surface is:!

And that the Earth orbital period is:!

We let the following ratio be approximated as 1:

So we have

We notice that . That is, the radius of the core of the Sun is on the order of the radius

of the Moon. the radius at the surface of the Sun, is the solar radius . We ﬁnd

must be multiplied by a factor somewhere between the golden ratio and 2/3 the ﬁbonacci

ratio that approximates it. We ﬁnd that value is . Thus our equation for the

temperature of the Sun is:!

This gives us!

From this we can compute the luminosity of the Sun:!

(

)

(

⊙

⋅

⊙

)

6α

⋅

4πh

⋅

(Ear th Day)

earth

⋅

(

)

⋅

6α

⋅

4πh

6α

⋅

4πh

(Ear th Day)

≈ 1

(

)

(

⊙

⋅

⊙

)

≈ Rm

⊙

= 16/25

(

⊙

)

(

⋅

⊙

)

(

⋅

1.7374E6

6.957E8

)

(

2E − 27

6.67E − 11

1.38E − 23

⋅

1.989E30

6.957E8

)(

5.972E 24

7.34767E 22

)

= 5,738

∘

⊙

= 4πR

⊙

σ T

of 48 79

= !

The actual solar luminosity is 3.846E-26 watts, so this is a very good estimate. We have our

system of equations for a Earth/Moon/Sun life bearing system is!

We can call these the Genesis Project equations after the Genesis project in the Star Trek movie

where they created a technology that converted dead worlds into living worlds."

4π(6.957E8)

(5.67E − 8)(5,738

∘

K )

= 3.738E 26wat ts

⊙

= 4πR

⊙

(

⊙

)

(

⋅

⊙

)

earth

⋅

(

)

⋅

6α

⋅

4πh

⊙

4π r

earth

= r

8π

1/2

= 67.9943kg

k =

(1 + α)

⋅ N

𝔼 =

773.5

⋅

k =

8π

(1 + α) ⋅ N

𝔼

k v

= 6

of 49 79

10.0 Another Theory For The Radius Of A Proton

The ancient Greeks knew motion to be an illusion. We see this in Zeno’s arrow. An arrow is ﬁred

so as to travel from A to B. Since to go to B it must travel to (1/2)B, then to half way to B from

there or to 1/4, then to halfway from 1/4 to B or to 1/8, and so on. Since we can keep dividing

the successive distances distances in half that must be traveled, the arrow never gets to B.

Mathematically this is!

As the number of partitions, , go to inﬁnity their widths go to zero and distances converges on

one. But for Zeno’s arrow the number of partitions never go to inﬁnity so their widths never go

to zero, so the distance never converges on one and we see that motion is an illusion. The

conundrum is where we have said . Is two raised to the zero really one? It was

established as a convention, there is no explanation for it. Couldn’t it just as easily be zero (two

times itself zero times is 0)? Or two times itself zero times is two? In!

∑

i=1

i = 1 + 2 + 3 + . . . + n =

n(n + 1)

lim

n⇀∞

= ∞

∑

i=1

(

)

(

)

(

)

(

)

+ . . . =

+ . . .

lim

n⇀∞

∑

i=1

(

)

(

)

(

)

(

)

+ . . . =

+ . . . = 1

lim

n⇀∞

n−1

∑

i=0

(

)

n(n + 1)

n−1

∑

i=0

(

)

= 2 −

n−1

lim

n⇀1

n−1

∑

i=0

(

)

= 2 −

1−1

= 2 −

= 1

of 50 79

we can say n can always take on a larger value so there can always be another

partition. If were zero then and any distance is inﬁnite, can never be traversed.!

In traveling from A to B something in motion at some time passes through all points but never

exists at any of them for any amount of time because there and inﬁnite number of points

between A and B and they have all been passed through in a ﬁnite amount of time. This is only

possible if the object in motion wasn’t at any point for any amount of time, or the journey would

have taken forever. Motion is adding up an inﬁnite amount of zeros adding up to a ﬁnite time.

Velocity is then the rate at which something passes through a point. Since an object in motion

exists at a point for no time regardless of its velocity, the time it is at one point is the same if its

velocity was diﬀerent. Something only exists at a given point for an amount of time if it has

stopped moving. So velocity must be described by energy, the energy it has at a point at a

given time. Thus we have Planck’s constant is or energy over time. Thus

energy is described by frequency:!

, , , , !

Then the rest frequency of a proton is !

10.1. !

10.2. !

We have!

10.3. !

If we take the radius of a proton to be!

0.8438E-15m which is very close to the CODATA results around 2018 (0.842E-15m) the we

have!

10.4. !

10.5. !

The result is that the radius of a proton is, given its mass!

10.6. !

10.7. , 10.8. !

n(n + 1)

= ∞

Joule ⋅ secon d s

E = h f

f = 1/s

h = J ⋅ s

h f = (J ⋅ s)(1/s)

E = J = Joules = energy

E = (1.67262E − 27kg)(299792459m /s)

= 1.503275897E − 10J

f =

1.503275897E − 10J

6.62607E − 34J ⋅ s

= 2.3687E 23s

−1

0.8438E − 15m

299,792,459m /s

= 2.8146E − 24s

(2.3687E 23s

−1

)(2.8146E − 24s) ≈ 0.6667 = 2/3

⋅

of 51 79

The radius of a proton is not well known, but are all well known.!

We found using aphelions and perihelions:!

Equation 1.6.

And we had equation 1.4

1.4.

Using 10.7 and 10.8

10.7. , 10.8.

And equation 1.4, we can describe the kinetic energies of the Moon and Earth with respect to

Earth rotation (Earth Day) in terms of proton mass and in terms of proton radius, as well as

come up for expressions for one second in terms of each of these.

10.9

10.10.

10.11.

10.12.

And dividing one by the other, we have

10.13.

That is

h, c, m

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

6α

h4π r

= 1.004996352secon d s

⋅

h 4π

= 1.062836secon d s

4α

⋅

h 4π

= 0.94536secon d s

⋅

h 4π

K E

(Ear th Day)

4α

⋅

h 4π

K E

(Ear th Day)

⋅

≈ 1

1.062836secon d s

0.94536secon d s

= 1.124

of 52 79

Which is about 89% accuracy. At this point we should write out what we had from the outset of

this paper combining equation 1.4 and 1.6!

Equation 10.14. !

And now we have!

Equation 10.11.

Equation 10.12.

And,…

Equation 1.4.

Equation 1.6.

And we now have!

10.9

10.10.

We see again that our Moon orbiting our Earth is pivotal to the Universe, and that the second is

a natural unit of measurement. If

10.13.

Is supposed to be exactly 1, then the radius of a proton is

10.15.

6α

h4π

(Ear th Day)

⋅

h 4π

K E

(Ear th Day)

4α

⋅

h 4π

K E

(Ear th Day)

6α

h 4π

= 1.0044996352

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

⋅

h 4π

= 1.062836secon d s

4α

⋅

h 4π

= 0.94536secon d s

⋅

≈ 1

⋅

(6.62607E − 34)

(1.67262E − 27)

⋅

299,792,459

= 0.880941E − 15m

of 53 79

11.0 Macro and Micro Planck Constants!

We found using aphelions and perihelions:!

Equation 1.6.

And we had equation 1.4

1.4. !

We write 1.6 in a form of Joule-seconds so we have a Planck constant given by the Earth orbit

which is determined by the Sun, we will call it the solar Planck constant which we will

distinguish from Planck’s constant by making the Planck constant the proton Planck

constant, . We have!

11. 1. !

11.2. !

From 1.4 we have!

11.3. !

From 10.6 we have!

10.6. !

11.4. !

These give!

11.5. !

From 11.4 we have!

11.6. !

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

6α

h4π r

= 1.004996352secon d s

⊙

earth

(1secon d ) = h

⊙

= J ⋅ s

1sec =

⊙

= (1sec)

(36α

)

4πr

⋅

⊙

⋅

⊙

⋅

of 54 79

= !

Using aphelions and perihelions!

Multiply that by our 1 second as a natural unit.!

If we use!

We can write equation 11.6 as!

11.7 !

This gives us that is given by some constant times the kinetic energy of the Earth. Let

us see what that constant is:!

11.8. !

= !

= This gives!

⊙

= (6.62607E − 34)

2.7396E33

299,792,459

⋅ 18769

⋅

(0.833E − 15)

1.67262E − 27)

⋅

(6.67408E − 11)

2.8314E33J ⋅ s

K E

earth

(5.972E 24kg)(30,290m /s)

= 2.7396E 33J

⊙

2.7396E33J

2.8314E33J

100 = 96.76 %

6α

h4π r

= 1.004996352secon d s

⊙

= (2.7396E 33J )(1.004966s) = 2.7533E 33J ⋅ s

⊙

2.7533E33J

2.8314E33J

100 = 97.24 %

⊙

⋅

2π r

⊙

C =

⋅

2π r

⋅

18769

299792459

2π (0.833E − 15

(6.67408E − 11)(1.67262E − 27)

1.5597656E33J

−1

≈ 1.56E 33J

−1

⊙

= h

(1.56E33J

−1

)KE

= (6.62607E − 34)(1.56E33)(2.7396E33)

of 55 79

We notice that !

So we have!

11.9. !

Where!

11.10. !

Which means is an Eigenvalue because it maps something into itself because !

11.12. !

Because!

Is approximately 1 second. In quantum mechanics eigenvalues describe atomic energy states,

discreet orbits of electrons around a proton. Here we see that is something happening but with

the Earth orbiting the Sun to relate the solar Planck constant on the macro-scale to the proton

Planck constant on the micro-scale. We can also say that!

11.13. !

Which makes an eigenvalue as well because it maps into because is the

macro-scale analog of . That is the incredible thing about !

Being approximately 1 second where 1 is the identity mapping something into itself. Thus we

have shown that the Earth/Sun/Moon orbital system is a quantum mechanical system

connected to the the proton/electron quantum mechanical systems. We include the Moon

because!

Is approximately 1 second. We have shown what we set out to do as well, and that is show that

the second is a natural unit, and a natural constant. We found it is

C = 1.033667s ≈ 1secon d

⊙

= (h

C )KE

C ) = 1secon d

⊙

= K E

6α

h4π r

= 1.004996352secon d s

⊙

= (C ⋅ K E

C ⋅ KE

⊙

6α

h4π r

= 1.004996352secon d s

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

of 56 79

11.14.

11.15. !

1secon d = h

C =

⋅

2π r

of 57 79

12.0 The Exact Radius of a Proton We may be able to come up with the actual value of the

radius of a proton by suggesting space and time are such that it is given by the golden ratio, .

Since we have 1 second is given by equation 1.4

Eq 1.4.

And we now have that one second is given by as well

Eq 12.1.

Then we have

Which is

Equation 12.2.

Which is our equation 10.15 for the radius of a proton

10.15.

We see this gives

Is an accuracy of 94.56%. We suggest that the nature of space and time is such that the equation

holds if is supposed to be . Then since , , and are well known, we

would be able to determine exactly what the the radius of a proton , is. Since

, we have

Equation 12.3.

This gives

Equation 12.4.

We see this is the momentum times distance of Quantum Mechanics. That is

6α

h 4π r

= 1.004996352secon d s

⋅

2π r

= 1.06secon d s

(

6α

⋅ r

4πh

)

3α

⋅

3Gm

2π r

= 1

⋅

⋅ m

= 1

⋅

299,792,459

6.62607E − 34

⋅

1.67262E − 27)(0.833E − 15)

= 0.945579897 = 1

3/2

Φ = ( 5 + 1)/2

Φ = 1/ϕ = ( 5 − 1)/2 = 0.618034

⋅

= r

= 0.816677E − 15m ≈ 0.817m

of 58 79

Equation 12.5.

Equation 12.6.

Equation 12.7.

We find it in the Schrödinger wave equation

Equation 12.8.

But we see as well this the the asymptote of set , the clock hyperbola, in Special Relativity in

spacetime diagrams, it is the world-line of a light signal. This is the hyperbola of special

relativity given by

Equation 12.9.

See following illustrations….

ϕh

= 1

= (mom ent u m)(dista nce) = px

ϕh = (m om ent u m)(dista n ce) = px

(iℏ)∂Ψ = − p∂xΨ

−

(

)

= 1

of 59 79

of 60 79

of 61 79

13.0 The Central Role of the Universe in the Emergence of Life The idea is to as I have

equation 1.2 and 1.3 for the hydrocarbons on page 7, I would like to next extend them to the

amino acids because they are they are the building block of life. That will take me to CHNOPS

the most abundant biological elements (Carbon, Hydrogen, Nitrogen, Oxygen, Phosphorus, and

Sulfur).

Equation 1.2.

Equation 1.3.

We found

Equation 13.1.

And, we can run our program to get more values by changing the value it increments for t, to get

approximate values for CHNOPS:

This one gives Sulfur S (16 protons) comes from t=0.375 seconds by incrementing t by 0.005

seconds. We can get the rest by incrementing t by 0.05 seconds:

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydr ogen(H )

proton s =

⋅

h 4π r

of 62 79

We get

Nitrogen N (7 protons) is t=0.8500 seconds

Oxygen O (8 protons) is t=0.7500 seconds

Phosphorus P (15 protons) is t=0.400 seconds

We already had for the hydrocarbon backbones to which N, O, P, S attach to make the amino

acids (they form what are called functional groups):

Hydrogen H (1 proton) is t=6.00 seconds

Carbon C (6 protons) is t=1.00 seconds

The interesting thing is carbons gives 1 second and hydrogen gives one second is the largest

integer before protons become fractional and the rest of the elements are at less than 1 second is

carbon. This shows the elements rooted in 6-fold symmetry based on hydrocarbons, the

backbones of life chemistry.

of 63 79

We see the rule of thumb is

Equation 13.2.

This is fascinating because we see this 6-fold symmetry predicts the protons in the elements in

terms of second because it would seem the second is a natural unit, as luck would have it,

because we have found

Equation 1.4

Equation 1.6.

KE is kinetic energy. The interesting thing is that we have the duration of a second is what it is

not because we were attempting to find some base unit of time for Nature, but rather we have

the duration of a second from ancient times that came to us over many years of dividing the

rotation of the Earth into hours, minutes, and seconds in such a way that we ended up with it.

However we may have had some sort of a thought going into it when we noticed

And that thought I showed (Section 3.0) was in a six-fold pattern that the motions of the Earth/

Moon/Sun system had in it approximately

Which can be written

Where there are 360 degrees in a circle and as such the Earth approximately moves through one

degree per day around the Sun.

Indeed we found that both the atom and the Earth/Moon/Sun orbital system quantize according

to a base unit of 1 second In section 11:

11.9.

11.10.

11.15.

Where is a Planck constant for an atom and is a Planck constant for the Sun.

proton s =

t im e(secon d s)

6α

h 4π r

= 1.004996352secon d s

K E

moon

K E

earth

(Ear th Day) = 1.08secon d s

1secon d =

365.25

⋅

= 0.0000000316881year

(

360d a ys

year

)(

24h ours

d a y

)(

60min

h our

)(

60sec

min

)

= 31104000secon d s /year

(

60d a ys

year

)(

1h our

d a y

)(

60min

h our

)(

60sec

min

)

⊙

= (h

C )KE

C ) = 1secon d

C =

⋅

2π r

G m

⊙

of 64 79

We indeed found this six-fold symmetry early on in the Periodic Table of the Elements by

forming an equation of state for it:

5.3

5.4

5.5

is any element and is the number of protons in times Avagadro’s number per that

number of grams. would be carbon is 6 grams per 6 protons. This resulted in an equation of

six-fold symmetry by way of a constant k, and the Earth orbital velocity:

5.1

We found that k is

Equation 13.3.

By way of finding an intermediate mass between the proton and the Chandrasekhar limit for a

white dwarf star:

5.8

5.7 Is the Chandrasekhar limit.!

5.9

By way of

5.11

Back to

Where time in seconds is given by

= 6E 23

proton s

gr a m

ℍ = 1

gr a m

proton

⋅ 𝔼 = 6E 23

𝔼

ℂ

k v

= 6

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

= Mm

M ≤ 0.77

ℏ

= 1.41 ⊙

ℏ

1/2

k =

(1 + α)

⋅ N

𝔼 =

773.5

⋅

proton s =

t im e(secon d s)

of 65 79

Carbon=6/1sec, Hydrogen=6/6sec, Nitrogen=6/0.85sec, Oxygen=6/0.75sec, and so on…

We want to pursue this to explain the amino acids, we have already found that the hydrocarbon

backbone of life chemistry is at the crux of explaining the 6-fold symmetry

Life chemistry as we know it is based on carbon. It is based on carbon because carbon is in

group 14 of the periodic table table which means it has 4 valence electrons to combine with other

atoms because since it is in group 18, the last group in the periodic table, the noble gases, has all

of its orbitals filled and it wants to have noble gas electron configuration, which means when it

bonds it gains 4 electrons; 18-14=4. Thus carbon gains 4 electrons when it combines with other

atoms, it becomes . We have said CHNOPS are the most abundant life elements. So for

instance to combine with hydrogen, which loses an electron to become it combines with 4 of

them to become neutral:

Which is methane, the most simple hydrocarbon. This relationship of 4 to 1 allows for the

backbone of life chemistry, the hydrocarbons. We have

proton s =

⋅

4π r

h 4π r

= 6proton ⋅ secon d s = carbon(C )

h 4π r

= 1proton ⋅ 6secon d s = hydr ogen(H )

4−

+ H

→ CH

of 66 79

But we said carbon is the basis for life because it ionizes , then why no silicon ? It ionizes

because it is in group 14 as well. The problem is silicon needs to combine with everything in

CHNOPS and it can’t but carbon can. The reason is that in the presence of oxygen gas, , it

combines with it very easily becoming , silicon dioxide, sand or the basic ingredient of

glass. We have

Silicon can gain or lose four electrons determining whether it attains the noble gas electron

configuration of either neon or argon. But silicon wants to gain or lose 4 electrons to get noble

has electron configuration, which is stable because all of the outer orbitals of a shell are filled.

Since we have already found the hydrocarbons in theory, the backbones of life chemistry, and

want to move on to the amino acids, the building block of life, we should talk a little about them.

The amino acids are the building block of life because deoxyribose nucleic acid (DNA) the

instructions for life use them to synthesize proteins, and proteins are used to make everything in

life, our hair, our nails, our tissue,…

To explain how life starts on Earth we need to explain how DNA and RNA (ribonucleic acid)

came into existence from what was present on the primordial Earth. As well we need to explain

how the amino acids came into existence. This has been done for some of the amino acids life

needs in the Miller-Urey experiment. Miller and Urey found if they put the primary hypothetical

constituents of the primordial Earth in a bottle and passed electricity through it, which were

methane , ammonia , and hydrogen gas , they were able to produce many of them.

Each amino acid has a hydrogen atom , a carboxyl group , and an amino group .

This is constant. But the variable attached to it, the so-called group, determines its function.

We have

4−

SiO

+ O

2−

+ O

2−

→ SiO

COOH

of 67 79

But we want to consider that (amino group) comes from (ammonia) of the primordial

Earth and that is really because is carboxylic acid and if we remove an it

becomes a carboxyl group. We have

I processed the 20 genetically encoded amino acids according to

Equation 13.4.

By molar mass which is to consider their relative masses and got

When I considered serine, which is

COOH

COO

COOH

Amin oGroup

Aci dGr oup

(Rgr oup)

N = 3(1.01) + 14.01 = 17.04g/m ol

COO = 12.01 + 2(16.00) = 44.01g /m ol

+ OH = 12.01 + 2.02 + 16.00 + 1.01 = 31.04g/m ol

of 68 79

I got

The molar mass of carbon is 12.01 g/mol.. So

Equation 13.5.

Thus the amino acid balances with the carbon, the core element of life, by mass. We notice that

Equation 13.6.

Where (phi) is the golden ratio conjugate is which occurs throughout life

because it has many physical functions.

Since we have

Where , we have

=1.0085862E-26

Thus

Which is what we already knew. But now we can find the time values fro the components of our

amino acids…

COO

(CH

+ OH ) =

17.04

44.01

(31.04) = 12.0182 ≈ C

COO

(CH

+ +OH ) = C

COO

= 0.387 ≈ (1 − ϕ)

ϕ =

5 − 1

h 4π r

= 1secon d

= carbon

h 4π r

= 18769

(6.62607E − 34)4π (0.833E − 15)

(6.67408E − 11)(299792459)

= carbon = 6(1.67262E − 27) = 1.00356E − 26

1.0085862E − 26

1.00356E − 26

= 1.005secon d s

N = 3(1.67262E − 27) + 7(1.67262E − 27) = 10(1.67262E − 27) = 1.67262E − 26

1.008562E − 26

1.67262E − 26

= 0.60298secon d s ≈ 0.603sec

COO = 22(1.67262E − 27) = 3.679764E − 26

1.008562E − 26

3.679764E − 26

= 0.27408sec ≈ 0.2741sec

of 69 79

This gives

We have for the time value of the R group

But let us find

Equation 13.7.

We have

This gives

We have already found by molar mass

Thus we have

Equation 13.8.

COO

0.603

0.2741

≈ 2.2

+ OH = 6 + 2 + * + 1 = 17

17(1.67262E − 27) = 2.843454E − 26

1.008562E − 26

2.843454E − 26

= 0.354696

COO

(CH

+ +OH ) =

0.603

0.2241

(0.354696sec) = 0.78sec

ser i ne =

COOH

(CH

+ OH )

N = 9(1.67262E − 27) = 1.505358E − 26

COOH = 23(1.67262E − 27) = 3.847E − 26

N =

1.008562E − 26

1.505358E − 26

= 0.66998sec

COOH =

1.008562E − 26

3.847E − 26

= 0.262168sec

COOH

0.262168

0.66998

= 0.3913 ≈ (1 − ϕ) = 0.382

COO

= 0.387 ≈ (1 − ϕ)

COOH

COO

of 70 79

Where the left side is by time units and the right side is by molar mass. The expression by molar

mass includes neutrons. We have

So the equation is the relationship between primordial Earth constituents and their products,

the amino acids. This is done seeing we found it was Carbon, the core element of life

This may shed some light on the prebiotic path to life. As it is figured in terms of the Universe

we see the Universe plays a central role in the emergence of life.

can be written , which was a dominant gas in the early Earth atmosphere, was as

well. In fact , oxygen, did not appear in abundance until the photosynthesizers arrived to

convert into , that is the plant life that makes carbon dioxide into oxygen. We can write

our equation as

Equation 13.9.

Equation 13.10.

We see the time value of carboxylic acid in an amino acid acting on carbon dioxide equals the

time value of the amino acid in an amino acid acting on ammonia. That is the primordial earth

gases ammonia and carbon dioxide correspond to amino acids (COOH and H2N).

Indeed we see the Universe plays a central role in the emergence of life. In the original paper I

presented a basic Universal energy that is obvious from dimensional analysis:

5.17

This resulted in the lifespan of the Universe which agrees with the Friedmann model of

the Universe (1E14 years). I found

COO

(CH

+ +OH ) = C

COO

COOH

)CO

= (t

)NH

⋅

= 1.599298E 29J

of 71 79

5.19

I found in terms of the orbit of the Moon around the Earth we get the Earth orbital period

5.21

I also found the radius of a proton is

Equation 13.11.

Equation 5.29

Using

Equation 5.25

Equation 13.12.

Remembering !

Noticing

Where is the Van Der Waals radius of a hydrogen atom (1.2E-10m). I also found it was

10.7. , 10.8.

There even became reason to suggest it was

Equation 12.3.

This gives

⋅

(

)

(1secon d ) = Li feSpa nUniverse

⋅

(

)

(1secon d ) = 1Ear thYear

⋅

4πh

= 8.288597 × 10

−16

m = 0.829f m

2π

⋅

𝔼

t =

2α

h 4π

k =

8π

(1 + α) ⋅ N

𝔼 =

773.5

(1 + α) ≈ 1

⋅

= r

of 72 79

Equation 12.4.

Which is toward what it seems to be approaching the more we try to measure (It is the least well

known of the data, but they feel they are closing in on it.)

This progress with finding the amino acids should open a door to a lot in looking for the origin

of life on Earth (a prebiotic path) because we now have

And we can substitute for 6 any number of protons, not just those in the amino acid serine as we

did, but in all of the genetically encoded amino acids that life utilizes. We can find the time

values as well for all of the elements of the periodic table. We can then proceed to all of the their

ensuing compounds which are billions upon billions. We can start with the millions of

fundamental biological compounds. But to do this we would need to write a computer program,

a much better one than the one I did on page 3. I think the next best step to take would be for

DNA and RNA and their phosphate backbones because to get started in the search for the

origins of life you need to start with DNA and RNA and the amino acids, to get the first living

cell.

Serine was the perfect amino acid to start with because its R group is simple using related

to the methane of the primordial Earth, and the hydroxyl ion so pivotal to the

function of water in life (the most important ingredient) that is the of dehydration

synthesis (removing water to make a bond) hydrolysis (decomposition by adding water to

remove a bond).

Finding Ribose

Several problems arise in trying to find a prebiotic path to life. We have to produce two things: 1.

The 20 genetically encoded amino acids for which the Miller-Urey chemistry only produces 11 of

them. 2. We have to produce the sugar ribose and the nucleobases (purines) adenine, guanine,

cytosine, thymine, and uracil so we can have the DNA and RNA that encodes for synthesizing

the amino acids into proteins.

Let us now move onto DNA and RNA for which biologists focus on RNA as the most likely

candidate for something to arrive first as a synthesizer of amino acids.

At the heart of solving this problem is the formose reaction. In this reaction the isomerizations

and aldol condensations seem to mimic the way current-day life makes sugars from smaller

precursors. It may be the most convenient chemistry by which sugars can be made. Problems

with it are that formaldehyde concentrations required to support the reaction can be difficult to

come by, though formaldehyde can be created by ultraviolet light on carbon dioxide and water in

the atmosphere that with the presence of carbon monoxide and water can produce from these

radicals formaldehyde, however only very slowly; it would require 10 million years for enough

formaldehyde to accumulate by this mechanism for the formose reaction to proceed. It is likely

in this way such a reaction could not be driven and the other problem is the formose reaction

produces only small amounts of each of many different sugars, and really only produces little

ribose, which is the most fundamental sugar to life on Earth. And worse, that little ribose

produced is very unstable. However, Steve Benner showed calcium borate which is in nature as

= 0.816677E − 15m ≈ 0.817m

h 4π r

= 1secon d

−

of 73 79

borax minerals and colemanite, stabilizes ribose under the conditions under which the formose

reaction can proceed. But, while this pushes up the levels of ribose, it does not stabilize the

ribose meaning it wouldn’t have the life span we would think would be needed for prebiotic

chemistry to make the precursors of life.

As far as producing purines of which Miller-Urey chemistry fails to produce in significant

amounts, the primary progress that has been made in this area was by John “Juan” Oro who

found that amino acids and small amounts of the purine base adenine form spontaneously from

mixtures of hydrogen cyanide and ammonia in water. However, higher concentrations of

hydrogen cyanide by hundreds of thousand of times are required than are calculated to have

existed on the surface of the early Earth. Hydrogen cyanide is more volatile than water so it

cannot be concentrated in tide pools. However, to that end progress was made by Leslie Orgel

who found, and was tested by by Miller in 1975, that it could be achieved by colder temperatures

as frozen hydrogen cyanide concentrated in the voids between ice crystals, but then the prebiotic

synthesis could have only happened at the Earth’s polar regions or in icy planetesimals in the

outer solar system. Miller did it at -78 deg centigrade.

Sugars are

Formaldehyde is . That is , but it is not a sugar, but all sugars are polymers of

formaldehyde.

Alexander Buterov discovered the formose reaction in 1861 which is the most plausible prebiotic

route path to sugars where prebiotic is prelife chemistry. It starts with formaldehyde dissolved

in water under alkaline basic conditions. At its peak 50% of the formaldehyde is converted to

sugars. Glycoaldehyde is formed from two formaldehyde molecules , then with another

formaldehyde makes the sugar glyceraldehyde with three carbons (n=3). This then isomerize to

dihydroxyacetone and again n=3 it just has a different geometry. These are the three possible

three carbon sugars. Two of these n=3 sugars can become hexoses (n=6 sugars). This

dihydroxyacetone (n=3) can combine with formaldehyde to form four carbon ketotetrose. We

can go to erythrulose which isomerize to aldotetrose, also four carbons. The formose reaction

produces little ribose (n=5) which is the most fundamental sugar for life.

The short of it is ribose ( ) of DNA and RNA is a polymer of formaldehyde ( ) so we

want to look at these two. Each atomic orbital of an atom can contain two electrons (an electron

pair) with opposite spins. However, in organic chemistry they occur briefly during a reaction on

an entity called a radical, or free radical. Thus a radical is an atom, molecule, or ion, that has at

least one unpaired valence electron, and these radicals are highly reactive which means they

have short lifetimes because they react to have their electrons paired. So a hydroxyl radical

would be where the dot means the oxygen has one unpaired electron and (triplet

carbene) has two unpaired electrons.

Back to the formose reaction. In order to have it we need formaldehyde. High energy electrons

or UV light cause the following radicals that strike atmospheric molecules knocking an electron

out making them:

n = 3,4,5,6,7

n = 1

HO·

: CH

→ 2H

⋅

of 74 79

And we get the following radical reactions with other atmospheric components:

Where is formaldehyde

Where is ethane.

We found in terms of time

And in terms of molar mass

Where and are primordial Earth substances and and are amino acid

components made up of their components. We want to do the same with the components of the

ribose sugar primary to life in DNA and RNA, which comes from formaldehyde. Therefore we

have since formaldehyde comes from

Molar Mass

Time Values

O → H

⋅

+ HO

⋅

→

⋅

+ H

O → H

C − O

⋅

+ H

C − O

⋅

+ H

⋅

→ H

C = O + H

C = O

⋅

→ H

C − CH

COOH

0.262168

0.66998

= 0.3913 ≈ (1 − ϕ) = 0.382

= 0.387 ≈ (1 − ϕ)

COOH

= 12.01 + 4(1.01) = 16.05g/m ol

= 2(1.01) = 2.02g /m ol

= 12.01 + 3(1.01) = 15.04g /m ol

O = 2(1.01) + 16..00 = 18.02g /m ol

CO = 2.02(1.01) + 12.01 + 16.00 = 30.03g /m ol

= 5(12.01) + 10 (1.01) + 8(16.00) = 198.15g /m ol

= (6 + 4) = 10(1.67262E − 27) = 1.67262E − 26

1.0085862E − 26

1.67262E − 26

= 0.603secon d s

of 75 79

By time values

We have

, ,

Which is

Which is:

, ,

= 2(1.67262E − 27) = 3.34524E − 27

1.0085862E − 26

3.34524E − 27

= 3.015secon d s

= (6 + 3)(1.67262E − 27) = 1.505E − 26

1.0085862E − 26

1.505E − 26

= 0.670secon d s

O = (2 + 8)(1.67262E − 27) = 1.67262E − 26

1.0085862E − 26

1.67262E − 26

= 0.603secon d s

CO) = (2 + 6 + 8)(1.67262E − 27) = 2.676192E − 26

1.0085862E − 26

2.6762E − 26

= 0.377secon d s

= (5 ⋅ 6 + 10 ⋅ 1 + 5 ⋅ 8) = 80(1.67262E − 27 ) = 1.3381E − 25

1.0085862E − 26

1.3381E − 25

= 0.07537secon d s

= 1.00

= 5.002

→ 2H

(m eth a n e) → CH

+ H

O → H

CO + H → H

CO( for m al dehyd e)

Pr im or dia l At m osper e → Fr eeRa dicals → for m al deyh yd e

→

→ H

0.603

3.015

= 0.2

0.670

0.603

= 1.1111

CO = 0.377secon d s ≈ (1 − ϕ)sec

of 76 79

We have for formaldeyhyde of which ribose is a polymer (using time values):

Equation 13.13.

Compare this to our equation for amino acids, it is similar in structure (using molar mass,

equation 13.5):

Equation 13.5.

Remember that we had by time value

And, by molar mass

So the factor by which everything works is because our result comes from

Often we see in Nature because if you go around a circle, is the same

location, just coming from the other direction.

DNA and RNA!

John Oro found small amounts of and the nucleobase adenine and amino acids when he mixed

hydrogen cyanide (HCN) and ammonia (NH3) in water (H2O). Later more experiments were

done and small amount of several diﬀerent nucleobases were found. Essentially what we had

were four molecules of HNC combine to make diaminomaleonitrile a polymer of hydrogen

cyanide which becomes . Then another molecule of HCN is added and it

becomes which is one of the four nucleases needed to make DNA, is adenine,

cytosine, guanine, thymine, or one of the four needed to make RNA is adenine, cytosine,

guanine, uracil.!

We have!

CO)

≈

1 − ϕ

⋅

h 4π

COO

(CH

+ OH ) = C

COOH

0.262168

0.66998

= 0.3913 ≈ (1 − ϕ) = 0.382

COO

= 0.387 ≈ (1 − ϕ)

(1 − ϕ)

6α

h4πr

= 1.004996352seconds

1 − ϕ

(HCN )

+ HCN → NH

= 14.01 + 3(1.01) = 17.04g /mol

of 77 79

The molar mass ratio of HCN to H2O is!

And by time values NH_3 to HCN is!

This gives us!

Equation 13.14 !

The time value of HCN is!

We have the molar mass of ammonium cyanide is!

And its time value is!

O = 2(1.01) + 16.00 = 18.02g/mol

= 7 + 3 = 10proton s

O = 2 + 8 = 10proton s

= 10(1.67262E − 27) = 1.67262E − 26

O = 10(1.67262E − 27) = 1.67262E − 26

1.0085862E − 27

1.67262E − 26

= 0.6029978sec ≈ 0.603secon d s

= 1.00

(HCN )

27.03g/m ol

18.02g/m ol

= 1.5

(NH

)

(HCN )

0.603s

0.402s

= 1.5

(NH

)

(HCN )

= (HCN )

HCN = (1 + 6 + 8)(1.67262E − 27) = 15(1.67262E − 27) = 2.50893E − 26

1.0085862E − 26

2.50893E − 26

= 0.4019985s ≈ 0.402secon ds

+ HCN → NH

CN = 14.01 + 4(1.01) + 12.01 + 14.01 = 44.07g /mol

of 78 79

We have for adenine!

The time value is!

Thus we see that!

Equation 13.15. !

Remember equation 13.14!

We have!

Equation 13.16. !

And this is our ﬁnal result. We see that adenine that encodes for life (its time value) to the time

value of ammonium cyanide times the molar mass of adenine over , the golden ratio, is the

ratio of the time values for the ammonia and cyanide times the molar mass of water which

make the ammonium cyanide on the left and when mixed in that water on the right make the

adenine on the left.!

CN = 1 + 4 + 6 + 7 = 24(1.67262 E − 27) = 4.014288E − 26

1.0085862E − 26

4.014288E − 26

= 0.25125s ≈ 0.25secon ds

= 5(12.01) + 5(1.01) + 5(14.01) = 135.15g /mol

= 5(6) + 5(1) + 5(7) = 70proton s

70(1.67262E − 27) = 1.170834E − 25

1.0085862E − 26

1.170834E − 25

= 0.086142545s ≈ 0.08614secon ds

CN = 0.25s

)

(HCN )

⋅

(NH

CN )

0.25

= 20s

−1

)

= 0.8164s

−1

(20s

−1

)(0.08164s

−1

) = 1.6328 ≈ = Φ = 1/ϕ

)

(HCN )

⋅

)

(NH

CN )

≈ Φ

(NH

)

(HCN )

= (HCN )

)

(NH

CN )

⋅

)

(NH

)

(HCN )

of 79 79

The Author!